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Alja [10]
3 years ago
9

Whats 3,684,266 to the nearest hundred thousand and million

Mathematics
1 answer:
grin007 [14]3 years ago
4 0
The nearest hundred thousand is 3,700,000
And the nearest million is 4,000,00
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AVERAGES 1. Find the mean, mode and median of: (a) 1.5, 1.2, 1.3, 1.5, 1.4 (b) 2.3, 1.2, 2.4, 1.2​
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Answer:

Look at Step by step

Step-by-step explanation:

Mean: (a) 1.38

Mean: (b) 1.7

median (a) 1.3

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Asthma is a condition in which it can be very hard for a person to breathe in enough air. Would it also be difficult for someone
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B, no, as long as the persons cells are getting different types of molecules from food he can still exercise.

Step-by-step explanation:

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Fruit basket is made with apple and oranges.the ratio of apples to oranges in a fruit basket is 2 to 3.if there are 45 pieces of
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1. Which angle is congruent to ∠1 ?
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An experiment to investigate the survival time in hours of an electronic component consists of placing the parts in a test cell
mixer [17]

Answer:

a) The sample mean is of 49 and the sample standard deviation is of 11.7.

b) The range of the true mean at 90% confidence level is of 9.62 hours.

c) The prediction interval, at a 90% confidence level, of it's failure time is between 39.38 hours and 58.62 hours.

Step-by-step explanation:

Question a:

Sample mean:

\overline{x} = \frac{34+40+46+49+61+64}{6} = 49

Sample standard deviation:

s = sqrt{\frac{(34-49)^2+(40-49)^2+(46-49)^2+(49-49)^2+(61-49)^2+(64-49)^2}{5}} = 11.7

The sample mean is of 49 and the sample standard deviation is of 11.7.

b)Determine the range of the true mean at 90% confidence level.

We have to find the margin of error of the confidence interval. Since we have the standard deviation for the sample, the t-distribution is used.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 6 - 1 = 5

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 5 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.9}{2} = 0.95. So we have T = 2.0.150

The margin of error is:

M = T\frac{s}{\sqrt{n}}

In which s is the standard deviation of the sample and n is the size of the sample. So

M = 2.0150\frac{11.7}{\sqrt{6}} = 9.62

The range of the true mean at 90% confidence level is of 9.62 hours.

(c)If a seventh sample is tested, what is the prediction interval (90% confidence level) of its failure time.

This is the confidence interval, so:

The lower end of the interval is the sample mean subtracted by M. So it is 49 - 9.62 = 39.38 hours.

The upper end of the interval is the sample mean added to M. So it is 49 + 9.62 = 58.62 hours.

The prediction interval, at a 90% confidence level, of it's failure time is between 39.38 hours and 58.62 hours.

4 0
2 years ago
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